Question

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A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.
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Answer 1

Length of the bar is ,l=2m

T1,T2 be the tensions produced in the left and right strings respectively.

At translational equilibrium, we have,

T1 sin36.9 =T2 sin53.1

$\frac{t1}{t2} = \frac{4}{3}$

For rotational equilibrium, on taking the torque about the centre of gravity,we have T1(cos36.9)×d=T2cos53.1 (2−d)

Using both equations,

d=0.72m

Hence, the centre of gravity of the given bar lies 0.72 m from its left end.

Answer 2

The free body diagram of the bar is shown in the following figure.

Length of the bar is ,l=2m

T 1,T 2 be the tensions produced in the left and right strings respectively.

At translational equilibrium, we have,

T 1 sin36.9 ∘ =T 2 sin53.1 .⟹ T 2T 1 = 34

For rotational equilibrium, on taking the torque about the centre of gravity, we have:

T 1 (cos36.9 ∘)×d=T 2cos53.1 ∘ (2−d)

Using both equations,

d=0.72m

Hence, the centre of gravity of the given bar lies 0.72 m from its left end.

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